1) I want to know the mechanism how to: show that the process $X_t$ solves this SDE
2) know if my friends though, mine though below are correct/incorrect.
I have the general linear stochastic diferential equation (SDE) with initial condition $X_0$, constants: $c,\sigma$ , I need to show that the process $X_t$ solves this SDE.
SDE:$$dX_t = c X_t dt + \sigma dW_t , t \in [0,T]$$
Stoch. Process:$$X_t = X_0 e^{ct}+\sigma e^{ct} \int_0^t e^{-cs}dW_s.$$
From simple logic, I think I need to insert $X_t$ to SDE( take differential of $X_t$ and receive exactly SDE I have above).
Is this mathematically correct way of showing that some solution solve SDE(or simple DE)?
My friend found in the web completely awkward solution to me:
1) Take differential by ?product formula?:
$$d(e^{-ct}X_t) = -c e^{-ct} X_t + e^{-ct} dX_t$$
2) Substitute our SDE $dX_t = c X_t dt + \sigma dW_t , t \in [0,T]$ to the second part of equation.
Finally integrate, and you get what you want:
$$X_t = …$$
My question: I think it is not correct proof, moreover, how come can I understand what should I differentiate ( I mean $d(e^{-ct}X_t)$ is not obvious). Is it? I think it somehow diminished proof.
I think that real proof is:
1) by ITOs formula:
$$df(t,W_t) = \dot f_t dt + \dot f_{W_t} dW_t + \frac{1}{2}\ddot f_{W_t,W_t} dt,$$
get $$d X_t = ..$$
2) see that it looks like $dX_t = c X_t dt + \sigma dW_t , t \in [0,T]$.
Here is how far I got, and what obstacles I have:
If I assume I can put $\frac{d}{dt}$ inside of integral and easily differentiate exponent $e^{-cs}$, I get:
$$dX_t = \frac{\partial }{\partial t} \left(X_0 e^{ct}+\sigma e^{ct} \int_0^t e^{-cs}dW_s \right) dt + \frac{\partial }{\partial W_t} \left(X_0 e^{ct}+\sigma e^{ct} \int_0^t e^{-cs}dW_s \right) dW_t + \frac{1}{2}\frac{\partial }{\partial W_t} \frac{\partial }{\partial W_t} \left(X_0 e^{ct}+\sigma e^{ct} \int_0^t e^{-cs}dW_s \right) dt, $$
or
$$dX_t = A dt + B dW_t + \frac{1}{2}C dt,$$
where:
$$ A= X_0 c e^{ct} + \left ( \frac{\partial }{\partial t} \sigma e^{ct} \int_0^t e^{-cs}dW_s \right) =$$
$$= X_0 c e^{ct} + \left ( \sigma ce^{ct} \int_0^t e^{-cs}dW_s + \sigma e^{ct} \int_0^t -c e^{-cs}dW_s \right).$$
Is it correct? Next,
$$B = 0 + \text{this is not easy for me to digest} = $$
$$ = 0 + \sigma e^{ct} e^{-ct} = \sigma$$
$$C = 0 + 0 \text{, because there is no } W_t$$
finally:
$$dX_t = \left ( X_0 c e^{ct} + \sigma ce^{ct} \int_0^t e^{-cs}dW_s + \sigma e^{ct} \int_0^t -c e^{-cs}dW_s \right) dt + \sigma dW_t + \frac{1}{2}0$$
$$dX_t = X_0 c e^{ct} dt+ \sigma dW_t $$
But here we see $X_0e^{ct}$, not $X_t$. =(
Best Answer
Your friends solution is correct. If $(X_t)_{t \geq 0}$ is a one-dimensional Itô process, then Itô's formula states $$ df(t,X_t)= \partial_x f(t,X_t) \, dX_t + \left(\frac{1}{2} \partial_x^2 f(t,X_t) \right) d\langle X \rangle_t + \partial_t f(t,X_t) \, dt. \tag{1}$$ Your friend used this identity for $f(t,x) := x e^{-ct}$.
Your attempt:
It this how you are taught to write down Itô's formula? In my oppinion, that's not a good way to write it this way. The problem is that you cannot apply Itô's formula this way. Itô's formula gives you the differential for $f(t,W_t)$ for (nice) functions $f$. But here, you want to calculate the differential of the expression
$$\int_0^t e^{-cs} \, dW_s,$$
i.e. we need a function $f$ such that
$$f(t,W_t) \stackrel{!??!}{=} \int_0^t e^{-cs} \, dW_s.$$
... tell me: How do you choose $f$? Before you have not chosen such a function $f$, you cannot apply Itô's formula this way. What you are doing is treating it as a constant and that's simply not correct.
In order to solve this SDE (or check that the given process is a solution to the SDE) you really have to use Itô's formula for Itô proceses, i.e. $(1)$.
Remark The solution your friend suggested applies Itô's formula to the process
$$e^{-ct} X_t \tag{1}$$
and, at the first glance, it is not obvious how to come up with this particular process. The idea is the following: Instead of considering the SDE
$$dX_t = c X_t \, dt + \sigma \, dW_t$$
we consider the corresponding ODE
$$dx_t = cx_t \, dt$$
(i.e. we just we leave away the stochastic part). It is well-known that the unique solution to this ordinary differential equation is given by
$$x_t = C e^{ct}$$
where $C \in \mathbb{R}$. So far, $C$ is some "deterministic" constant. Now, however, we return to our stochastic setting and allow $C$ to depend on $\omega$ (this is the counterpart of the variation of constants-approach for SDEs). So, by the previous identity, our new auxilary process $C$ is given by
$$C_t = e^{-ct} X_t$$
... and this is exactly the process from $(1)$.
There are a lot of examples where this approach [i.e. first solve the corresponding ODE and then make a "stochastic" variation of constants] works, ee e.g. this question. However, I don't know any statements for which types of SDEs this approach works and for which it doesn't.